CirclesClass 10 Mathematics NCERT Solutions

Answer: A circle can have infinitely many tangents.

Reason: A circle is a collection of infinitely many points. A unique tangent can be drawn at each of these points. Therefore, a circle can have an infinite number of tangents.

Answer: (D) $\sqrt{119}$ cm.

Explanation: Given: Radius of the circle, $OP = 5 \text{ cm}$. Distance of point Q from the center, $OQ = 12 \text{ cm}$. PQ is the tangent to the circle at point P.

To Find: The length of the tangent, PQ.

Solution: According to Theorem 10.1, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $\angle OPQ = 90^\circ$. This means that $\triangle OPQ$ is a right-angled triangle with the right angle at P.

By the Pythagorean theorem: $OQ^2 = OP^2 + PQ^2$ $12^2 = 5^2 + PQ^2$ $144 = 25 + PQ^2$ $PQ^2 = 144 - 25$ $PQ^2 = 119$ $PQ = \sqrt{119} \text{ cm}$

Final Answer: The length of PQ is $\sqrt{119}$ cm, which corresponds to option (D).

To Construct: A circle, a given line, a tangent parallel to the given line, and a secant parallel to the given line.

Result: We have a circle, a given line L, a tangent parallel to L, and a secant parallel to L.

Answer: (A) 7 cm

Explanation: Given: Let the circle have its center at O. Let Q be the external point. Length of the tangent from Q, let us say QT, is $QT = 24 \text{ cm}$. Distance of Q from the center, $OQ = 25 \text{ cm}$.

To Find: The radius of the circle, $OT$.

Solution: The radius at the point of contact is perpendicular to the tangent. So, $\triangle OTQ$ is a right-angled triangle with the right angle at T.

By the Pythagorean theorem: $OQ^2 = OT^2 + QT^2$ $25^2 = OT^2 + 24^2$ $625 = OT^2 + 576$ $OT^2 = 625 - 576$ $OT^2 = 49$ $OT = \sqrt{49} = 7 \text{ cm}$

Final Answer: The radius of the circle is 7 cm, which corresponds to option (A).

Answer: (B) $70^\circ$

Explanation: Given: TP and TQ are tangents to a circle with center O. $\angle POQ = 110^\circ$.

To Find: $\angle PTQ$.

Solution: We know that the radius is perpendicular to the tangent at the point of contact. Therefore, $OP \perp TP$, which means $\angle OPT = 90^\circ$. And, $OQ \perp TQ$, which means $\angle OQT = 90^\circ$.

Now, consider the quadrilateral TPOQ. The sum of all interior angles of a quadrilateral is $360^\circ$. $\angle OPT + \angle POQ + \angle OQT + \angle PTQ = 360^\circ$ $90^\circ + 110^\circ + 90^\circ + \angle PTQ = 360^\circ$ $290^\circ + \angle PTQ = 360^\circ$ $\angle PTQ = 360^\circ - 290^\circ$ $\angle PTQ = 70^\circ$

Final Answer: $\angle PTQ$ is equal to $70^\circ$, which corresponds to option (B).

Answer: (A) $50^\circ$

Explanation: Given: PA and PB are tangents from point P to a circle with center O. The angle between the tangents, $\angle APB = 80^\circ$.

To Find: $\angle POA$.

Solution: In triangles $\triangle OAP$ and $\triangle OBP$: $OA = OB$ (Radii of the same circle) $PA = PB$ (Lengths of tangents from an external point are equal) $OP = OP$ (Common side)

By SSS congruence rule, $\triangle OAP \cong \triangle OBP$. Therefore, $\angle OPA = \angle OPB = \frac{1}{2} \angle APB = \frac{1}{2}(80^\circ) = 40^\circ$. Also, $\angle OAP = \angle OBP = 90^\circ$ (Radius is perpendicular to the tangent at the point of contact).

Now, in $\triangle OAP$, the sum of angles is $180^\circ$. $\angle OAP + \angle OPA + \angle POA = 180^\circ$ $90^\circ + 40^\circ + \angle POA = 180^\circ$ $130^\circ + \angle POA = 180^\circ$ $\angle POA = 180^\circ - 130^\circ$ $\angle POA = 50^\circ$

Final Answer: $\angle POA$ is equal to $50^\circ$, which corresponds to option (A).

To Prove: The tangents drawn at the ends of a diameter of a circle are parallel.

Proof: Let us consider a circle with center O and diameter AB. Let PQ be the tangent to the circle at point A, and let RS be the tangent to the circle at point B.

We know that the radius is perpendicular to the tangent at the point of contact. So, $OA \perp PQ$. This implies $\angle OAP = 90^\circ$. Similarly, $OB \perp RS$. This implies $\angle OBS = 90^\circ$.

Now, AB is a straight line (the diameter), which acts as a transversal for the lines PQ and RS. We have $\angle PAB = \angle OAP = 90^\circ$. We also have $\angle ABS = \angle OBS = 90^\circ$.

So, $\angle PAB = \angle ABS = 90^\circ$. These two angles are a pair of alternate interior angles. When a transversal intersects two lines such that the alternate interior angles are equal, the lines are parallel.

Therefore, $PQ \parallel RS$.

To Prove: The perpendicular at the point of contact to the tangent to a circle passes through the centre.

Proof: Let us consider a circle with center O. Let XY be a tangent to the circle at a point P. We need to prove that the line perpendicular to XY at P passes through the center O.

We know from Theorem 10.1 that the radius from the center to the point of contact is perpendicular to the tangent. So, $OP \perp XY$, which means $\angle OPX = 90^\circ$.

Now, let us assume there is another line, say ZP, which is perpendicular to XY at point P, but it does not pass through the center O. If $ZP \perp XY$, then $\angle ZPX = 90^\circ$.

From our findings, we have $\angle OPX = 90^\circ$ and $\angle ZPX = 90^\circ$. This implies $\angle OPX = \angle ZPX$. This is only possible if the line segments OP and ZP are collinear, meaning the line ZP must pass through O. This contradicts our assumption that ZP does not pass through O.

Therefore, our assumption was incorrect. The perpendicular at the point of contact to the tangent must pass through the center of the circle.

Given: Let the circle have its center at O. Point A is at a distance of 5 cm from the center, so $OA = 5 \text{ cm}$. The length of the tangent from point A is 4 cm. Let P be the point of contact, so $AP = 4 \text{ cm}$.

To Find: The radius of the circle, $OP$.

Solution: The radius is perpendicular to the tangent at the point of contact. So, $OP \perp AP$, which means $\angle OPA = 90^\circ$. $\triangle OPA$ is a right-angled triangle.

By the Pythagorean theorem: $OA^2 = OP^2 + AP^2$ $5^2 = OP^2 + 4^2$ $25 = OP^2 + 16$ $OP^2 = 25 - 16$ $OP^2 = 9$ $OP = \sqrt{9} = 3 \text{ cm}$

Final Answer: The radius of the circle is 3 cm.

Given: Two concentric circles with a common center O. Radius of the larger circle, $R = 5 \text{ cm}$. Radius of the smaller circle, $r = 3 \text{ cm}$. AB is a chord of the larger circle that is a tangent to the smaller circle at point P.

To Find: The length of the chord AB.

Solution: Join OA and OP. OA is the radius of the larger circle, so $OA = 5 \text{ cm}$. OP is the radius of the smaller circle, so $OP = 3 \text{ cm}$. Since AB is a tangent to the smaller circle at point P, the radius OP is perpendicular to the chord AB. Thus, $\angle OPA = 90^\circ$.

Now, consider the right-angled triangle $\triangle OPA$. By the Pythagorean theorem: $OA^2 = OP^2 + AP^2$ $5^2 = 3^2 + AP^2$ $25 = 9 + AP^2$ $AP^2 = 25 - 9 = 16$ $AP = \sqrt{16} = 4 \text{ cm}$

We also know that a perpendicular drawn from the center of a circle to a chord bisects the chord. Since $OP \perp AB$, P is the midpoint of AB. Therefore, $AB = 2 \times AP$. $AB = 2 \times 4 = 8 \text{ cm}$

Final Answer: The length of the chord of the larger circle is 8 cm.

To Prove: For a quadrilateral ABCD circumscribing a circle, $AB + CD = AD + BC$.

Proof: Let the quadrilateral ABCD touch the circle at points P, Q, R, and S on the sides AB, BC, CD, and DA, respectively.

We know that the lengths of tangents drawn from an external point to a circle are equal. Therefore, from each vertex of the quadrilateral, we have:

Now, let us add these four equations: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$

From the figure, we can see that: $AP + BP = AB$ $CR + DR = CD$ $AS + DS = AD$ $BQ + CQ = BC$

Substituting these into the summed equation: $AB + CD = AD + BC$

To Prove: $\angle AOB = 90^\circ$.

Proof: Let the circle have center O. Let XY be a tangent at point P and X'Y' be a tangent at point Q. Since XY is parallel to X'Y', PQ must be a diameter of the circle. Let AB be another tangent to the circle at point C.

Join OA, OB, and OC.

Consider triangles $\triangle OPA$ and $\triangle OCA$. $OP = OC$ (Radii of the same circle) $OA = OA$ (Common side) $AP = AC$ (Tangents from an external point A are equal in length)

By SSS congruence rule, $\triangle OPA \cong \triangle OCA$. Therefore, $\angle POA = \angle COA$. ... (1)

Similarly, consider triangles $\triangle OQB$ and $\triangle OCB$. $OQ = OC$ (Radii of the same circle) $OB = OB$ (Common side) $BQ = BC$ (Tangents from an external point B are equal in length)

By SSS congruence rule, $\triangle OQB \cong \triangle OCB$. Therefore, $\angle QOB = \angle COB$. ... (2)

Since PQ is a diameter, it is a straight line. Therefore, the angle $\angle POQ$ is $180^\circ$. $\angle POQ = \angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ$ From (1) and (2), we can write: $\angle COA + \angle COA + \angle COB + \angle COB = 180^\circ$ $2\angle COA + 2\angle COB = 180^\circ$ $2(\angle COA + \angle COB) = 180^\circ$ $\angle COA + \angle COB = 90^\circ$

From the figure, we can see that $\angle AOB = \angle COA + \angle COB$. Therefore, $\angle AOB = 90^\circ$.